Sound Waves

Q: What is Laplace's correction to Newton's formula?

Newton treated sound in a gas as isothermal and got v = root of P over rho, about 280 m/s in air. Laplace showed the compressions and rarefactions are too fast for heat to flow, so the process is adiabatic, the bulk modulus is gamma P, and v = root of gamma P over rho, about 331 m/s, matching experiment.

PhysicsWavesFor NEET aspirants

Sound waves are mechanical, longitudinal waves produced by a vibrating source and carried by a medium as a series of compressions and rarefactions. The speed of sound is $v = E / ρ$ for the medium: $Y / ρ$ in a solid rod and $γ P / ρ = γ R T / M$ in a gas (Laplace), about $331 m s^{- 1}$ in air at $0^{\circ} C$ . This page covers how sound waves are produced, their speed and what changes it, pitch, loudness in decibels, quality and echo. Tested every year in JEE Main and NEET.

On this page1Production2Audible range3Speed in solids and fluids4Newton's formula5Laplace correction6Temperature, pressure, humidity, wind7Pitch, loudness, quality8Echo

Key Formulas - Quick Reference

★ Must learnSpeed of longitudinal waves: solid rod $v = \frac{Y}{ρ}$ ; fluid $v = \frac{B}{ρ}$ , with $B = - V \frac{d P}{d V}$
Newton (isothermal): $B = P$ , $v = \frac{P}{ρ} \approx 280 m s^{- 1}$ in air at NTP (too low)
★ Must learnLaplace (adiabatic): $B = γ P$ , $v = \frac{γ P}{ρ} = \frac{γ R T}{M} \approx 331 m s^{- 1}$ in air at NTP
★ Must learnTemperature: $v \propto T$ , $\frac{v _{1}}{v _{2}} = \frac{T _{1}}{T _{2}}$ ; $v_{t} \approx v_{0} + 0.61 t$ ( $t$ in $^{\circ} C$ )
Pressure: no effect at constant $T$ ( $P / ρ$ constant). Humidity: raises $v$ . Wind: $v_{eff} = v + w cos θ$
Different gases at the same $T$ : $v \propto \frac{γ}{M}$
★ Must learnSound level: $β = 10 lo g_{10} \frac{I}{I _{0}}$ dB, $I_{0} = 1 0^{- 12} W m^{- 2}$ ; $β_{2} - β_{1} = 10 lo g_{10} \frac{I _{2}}{I _{1}}$
Point source: $I = \frac{P}{4 π r ^{2}}$ , so $I \propto \frac{1}{r ^{2}}$
Audible range $20 Hz$ to $20 kHz$ ; echo needs $t \geq 0.1 s$ , i.e. $d \geq 17 m$ in air

1. Sound: Production and Nature

Sound is produced by a vibrating body. Strike the gong of a bell with a hammer and it vibrates; touch it and you can feel the vibrations. The vibrations pass to the air, travel through the air to the ear, set the eardrum vibrating, and these vibrations are passed on to the brain, which we perceive as sound. A guitar string, the vocal cords, the prongs of a tuning fork and the diaphragm of a loudspeaker are all such sources.

★ Must learnSound is a mechanical, three-dimensional, longitudinal wave. Being mechanical, it needs a medium with inertia and elasticity, and it travels through the medium as a series of periodic compressions (pressure slightly above normal) and rarefactions (pressure slightly below normal) produced by the vibrating source.

1.1 How a tuning fork sends out sound

At rest, the air in front of the prongs is at normal atmospheric pressure; the air layers are evenly spaced.
When prong $B$ moves outwards, it pushes the air in front of it, and the pressure there rises slightly. This region of higher pressure is a compression pulse; it travels away from the prong at the speed of sound.
Prong $B$ then reverses and moves inwards. It drags air away from the region in front of it, and the pressure dips slightly below normal. This region is a rarefaction pulse, and it follows immediately behind the compression, also at the speed of sound.
Each vibration repeats the pair, so a continuous train of compressions and rarefactions, a sound wave of the fork's frequency, spreads outwards.

Figure 1: Each outward swing of the prong launches a compression, each inward swing a rarefaction. A fork vibrating at

f

sends out

f

compression-rarefaction pairs per second: a sound wave of frequency

f

Sound needs a medium: the bell-jar experiment. An electric bell ringing inside a glass jar is heard clearly. As a vacuum pump removes the air, the sound grows fainter and almost stops, although the hammer can still be seen striking the gong. With air let back in, the sound returns. On the Moon, astronauts talk by radio for the same reason.

Figure 2: Bell-jar experiment. (a) With air in the jar the ringing is heard. (b) As the pump removes the air the sound fades almost completely, though the hammer is still seen striking the gong: sound needs a material medium, light does not.

Key idea

The air does not travel from the fork to your ear. Each layer only shuffles back and forth; the pattern of compressions and rarefactions travels.

2. Audible, Infrasonic and Ultrasonic Sound

A healthy human ear responds to longitudinal waves of frequency from about $20 Hz$ to $20 kHz$ ; this band is audible sound. The upper limit falls with age.

Range	Frequency	Examples and uses
Infrasonic	below $20 Hz$	Earthquakes, volcanoes, whales and elephants communicating over long distances
Audible	$20 Hz$ to $20 kHz$	Speech, music; human speech is mostly $100 Hz$ to $4 kHz$
Ultrasonic	above $20 kHz$	Bats and dolphins (echolocation), SONAR, cleaning, medical scans at a few $MHz$

Figure 3: Humans hear roughly

20 Hz

20 kHz

. Infrasound lies below, ultrasound above (bats, SONAR, medical imaging at a few

MHz

3. Speed of Longitudinal Waves in Solids and Fluids

3.1 Derivation for a solid rod

Consider a long rod of cross-sectional area $S$ and density $ρ$ , with the source at $O$ . A slice $A B$ has its face $A$ at distance $x$ and its face $B$ at $x + d x$ . As the wave passes, the particles at $A$ are displaced by $y$ and those at $B$ by $y + d y$ , so the slice is stretched by $d y$ .

Figure 4: The slice

A B

of width

d x

is displaced by

y

and stretched by

d y

. Strain

= \frac{d y}{d x}

; the unbalanced force

d F

accelerates the slice. This gives

v = Y / ρ

Stress at a cross-section: $\frac{F}{S}$ . Strain in the slice: $\frac{d y}{d x}$ (change in length over original length).
Young's modulus: $Y = \frac{F / S}{d y / d x}$ , so $F = Y S \frac{\partial y}{\partial x}$ .
The net force on the slice is $d F = \frac{\partial F}{\partial x} d x = Y S \frac{\partial ^{2} y}{\partial x ^{2}} d x$ .
Newton's second law for the slice of mass $d m = ρS d x$ : $d F = (ρS d x) \frac{\partial ^{2} y}{\partial t ^{2}}$ .
Equate the two:
$\frac{\partial ^{2} y}{\partial t ^{2}} = \frac{Y}{ρ} \frac{\partial ^{2} y}{\partial x ^{2}}$
Compare with the wave equation $\frac{\partial ^{2} y}{\partial t ^{2}} = v^{2} \frac{\partial ^{2} y}{\partial x ^{2}}$ :
$v = \frac{Y}{ρ}$

3.2 Liquids and gases

A fluid has no fixed length to stretch; it responds to squeezing through its bulk modulus. Repeating the argument with volume strain in place of length strain gives

v = \frac{B}{ρ}, B = - \frac{d P}{d V / V} = - V \frac{d P}{d V}

The minus sign makes

B

positive, because volume falls when pressure rises. For a gas,

B

depends on how the gas is compressed, which is exactly where Newton and Laplace differ.

Figure 5: Speed of sound in

m s^{- 1}

(NCERT values). Sound is fastest in solids, slower in liquids and slowest in gases; among gases, lighter gases carry it faster.

Why solids are fastest: solids are only a few times denser than liquids, but their elastic moduli are far larger (steel: $Y \approx 2 \times 1 0^{11} Pa$ ; water: $B \approx 2.2 \times 1 0^{9} Pa$ ; air: $B \approx 1.4 \times 1 0^{5} Pa$ ). The modulus wins, so $v_{solid} > v_{liquid} > v_{gas}$ .

4. Newton's Formula for the Speed of Sound in a Gas

Newton assumed that the temperature of the gas stays constant as sound passes (an isothermal process), so Boyle's law applies to each layer.

Isothermal: $P V = constant$ . Differentiate: $P d V + V d P = 0$ .
So $- V \frac{d P}{d V} = P$ , that is $B_{iso} = P$ : the bulk modulus equals the pressure.
Hence
$v = \frac{P}{ρ}$
and with the gas law $P = \frac{ρR T}{M}$ , $v = \frac{R T}{M}$ .
Air at NTP: $P = 1.01 \times 1 0^{5} Pa$ , $ρ = 1.293 kg m^{- 3}$ :
$v = \frac{1.01 \times 1 0 ^{5}}{1.293} \approx 279.5 m s^{- 1}$

Experiments give about $331$ to $332 m s^{- 1}$ at NTP. A gap of over $50 m s^{- 1}$ (about $16%$ ) is far too large to be experimental error. Newton could not explain it; Laplace did.

5. Laplace's Correction

Laplace pointed out that compressions and rarefactions follow each other very quickly, and air is a poor conductor of heat. Where the gas is compressed it warms up; where it is rarefied it cools. There is no time for heat to flow between them, so the temperature is not constant; instead no heat enters or leaves each layer. Sound propagation in a gas is an adiabatic process.

Adiabatic: $P V^{γ} = constant$ , with $γ = \frac{C _{p}}{C _{v}}$ .
Differentiate: $γ P V^{γ - 1} d V + V^{γ} d P = 0$ , so $- V \frac{d P}{d V} = γ P$ .
The bulk modulus is now $B_{adia} = γ P$ , and
$v = \frac{γ P}{ρ} = \frac{γ R T}{M}$
Air ( $γ = 1.40$ ) at NTP:
$v = \frac{1.40 \times 1.01 \times 1 0 ^{5}}{1.293} \approx 330.7 m s^{- 1}$
in agreement with experiment.

Figure 6: The adiabatic curve is

γ

times steeper, so air is stiffer to fast compressions:

B = γ P

(Laplace) instead of

B = P

(Newton). Speed rises by

γ \approx 1.18

★ Must learnLaplace's formula:

v = \frac{γ P}{ρ} = \frac{γ R T}{M}

. It is Newton's value multiplied by

γ

331 \approx 279.5 \times 1.40

. Values of

γ

: monatomic gas

5/3

, diatomic gas (air,

O_{2}

N_{2}

)

7/5

, polyatomic about

4/3

Exam Trick

Compare gases in one line. At the same temperature, $v \propto γ / M$ . Hydrogen and oxygen (both diatomic): $\frac{v _{H_{2}}}{v _{O_{2}}} = \frac{32}{2} = 4$ . Helium versus air: $\frac{( 5/3 ) /4}{1.4/29} \approx 2.9$ . No need to know $P$ or $ρ$ .

6. Factors Affecting the Speed of Sound in a Gas

6.1 Temperature

For a given gas, $R$ , $M$ and $γ$ are fixed, so $v \propto T$ with $T$ in kelvin:

\frac{v _{1}}{v _{2}} = \frac{T _{1}}{T _{2}}

At $t^{\circ} C$ : $v_{t} = \frac{γ R ( 273 + t )}{M} = v_{0} (1 + \frac{t}{273})^{1/2}$ . For small $t$ , the binomial approximation gives

v_{t} \approx v_{0} (1 + \frac{t}{546})

so the speed rises by $\frac{v _{0}}{546} = \frac{331}{546} \approx 0.61 m s^{- 1}$ for every $1^{\circ} C$ .

Figure 7:

v \propto T

for a given gas. Near room temperature the curve is almost straight, which is why

v_{t} \approx v_{0} + 0.61 t

works for small

t

(in

^{\circ} C

JEE Advanced

Sound speed and molecular speed. Compare $v = \frac{γ R T}{M}$ with the rms speed of the molecules, $v_{rms} = \frac{3 R T}{M}$ :

\frac{v}{v _{rms}} = \frac{γ}{3}

about $0.68$ for air. Sound is passed on by molecular collisions, so it is always slower than the molecules themselves. For a mixture of gases use $M_{mix} = \frac{n _{1} M _{1} + n _{2} M _{2}}{n _{1} + n _{2}}$ and find $γ_{mix}$ from $C_{v, mix} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}}$ .

6.2 Pressure

From the gas law, $\frac{P}{ρ} = \frac{R T}{M}$ . At constant temperature, if the pressure changes, the density changes in the same proportion, so $P / ρ$ stays the same and $v = γ P / ρ$ does not change. The speed of sound in a gas is independent of pressure (at constant temperature). This is why sound travels at the same speed on a hill and at sea level on days of equal temperature.

6.3 Humidity

Water vapour (density about $0.8 kg m^{- 3}$ at NTP) is lighter than dry air ( $1.293 kg m^{- 3}$ ). Moist air at the same pressure therefore has a lower density, so $v_{moist air} > v_{dry air}$ . Sound travels slightly faster on a humid day.

6.4 Wind

Wind carries the medium along. If the wind has velocity $w$ and sound travels at $v$ relative to still air, the velocity relative to the ground is $v + w$ . Along the line of travel,

v_{eff} = v + w cos θ

where $θ$ is the angle between the wind and the direction of the sound: $v + w$ with the wind, $v - w$ against it.

Factor	Effect on $v$ in a gas	Reason
Temperature $T$ rises	Increases, $v \propto T$	$v = γ R T / M$
Pressure changes ( $T$ fixed)	No change	$P / ρ$ stays constant
Humidity rises	Increases slightly	Moist air is less dense
Molar mass $M$ larger	Decreases, $v \propto 1/ M$	Heavier molecules, more inertia
$γ$ larger	Increases, $v \propto γ$	Stiffer under adiabatic compression
Wind along the sound	Increases, $v + w cos θ$	The medium itself moves
Frequency or wavelength	No change	Speed depends on the medium only

Quick Recall: tap to check

Why did Newton's formula give too low a value?

He took the process as isothermal (

B = P

); it is adiabatic (

B = γ P

), so his value was

γ

times too small.

At what temperature is the speed of sound in air double its value at

0^{\circ} C

T = 4 \times 273 = 1092 K

, i.e.

819^{\circ} C

Does sound travel faster on a mountain top (lower pressure, same temperature)?

No: at the same temperature the speed does not depend on pressure.

Figure 8: Pick the elastic modulus that matches the medium:

Y

for a rod,

B

for a liquid,

γ P

for a gas. Then only temperature, humidity and the kind of gas change

v

in a gas.

7. Characteristics of Sound: Pitch, Loudness and Quality

Our ear describes any sound by three characteristics. Each one is linked to a physical property of the wave.

Sensation (what we hear)	Physical property	Relation
Pitch: shrill or grave	Frequency	Higher frequency, higher pitch
Loudness: loud or soft	Intensity (energy per second per unit area)	Roughly logarithmic: measured in decibels
Quality or timbre	Waveform (mix of overtones)	Same note, different shape

7.1 Pitch and frequency

Pitch is the sensation by which we tell a buffalo's voice (low pitch) from a man's (higher) and a woman's or a child's (still higher). It depends mainly on the dominant frequency in the sound: the higher the frequency, the higher the pitch. A shrill whistle has a high pitch; a drumbeat has a low one.

7.2 Loudness, intensity and the decibel

Loudness is related to intensity, but not in direct proportion: a sound ten times more intense does not seem ten times louder. Our sense of loudness follows the logarithm of intensity, so we measure sound level in decibels:

β = 10 lo g_{10} \frac{I}{I _{0}} dB, I_{0} = 1 0^{- 12} W m^{- 2}

I_{0}

is roughly the faintest audible intensity at middle frequencies; at

I = I_{0}

β = 0

. Two levels differ by

β_{2} - β_{1} = 10 lo g_{10} \frac{I _{2}}{I _{1}}

Figure 9:

β = 10 lo g_{10} (I / I_{0})

. Every

+ 10 dB

multiplies the intensity by 10; every

+ 20 dB

by 100. Levels above about

85 dB

(red) harm hearing over time.

For a small source radiating power $P$ equally in all directions, the energy spreads over a sphere, so $I = \frac{P}{4 π r ^{2}}$ : doubling the distance cuts the intensity to one quarter, a drop of about $6 dB$ .

Exam Trick

Decibels add by powers of ten. $\times 10$ in intensity is $+ 10 dB$ ; $\times 100$ is $+ 20 dB$ ; $\times 2$ is $+ 3 dB$ (since $lo g_{10} 2 \approx 0.3$ ). So two identical $70 dB$ machines together give $73 dB$ , not $140 dB$ . Halving the distance from a point source gives $+ 6 dB$ .

7.3 Quality and waveform

A source seldom produces one pure frequency. Along with the fundamental, it produces weaker higher frequencies (overtones) with different amplitudes, and their superposition gives the actual waveform. Two instruments playing the same note (the same fundamental, say $440 Hz$ ) at equal loudness still sound different because their waveforms differ. This is quality (timbre): we tell a tabla from a mridang, or recognise a friend's voice on the phone, by it.

Figure 10: Same fundamental (same

T

, same pitch), different overtones mixed in. The waveform shape is what the ear hears as quality or timbre.

Loudness vs intensity

Intensity is a physical quantity, in $W m^{- 2}$ , measured by instruments. Loudness is a sensation; it grows with intensity roughly logarithmically and also depends on the listener's ear and on frequency.

Pitch vs frequency

Frequency is physical, in Hz. Pitch is how high or low we hear it. They rise together, but pitch is a sensation and cannot be measured with a meter.

8. Echo

The repetition of a sound caused by reflection from a distant, large surface such as a cliff, a hill, a well or a building is called an echo. The sensation of a sound persists in our ear for about $\frac{1}{10} s$ . If the reflected sound returns in less time than this, it merges with the original and no separate echo is heard. The same limit is why the ear cannot follow beats faster than about $10$ per second (see the Beats concept).

Figure 11: Echo time

t = \frac{2 d}{v}

. The sensation of a sound lasts about

0.1 s

, so a distinct echo needs

d \geq \frac{v \times 0.1}{2} \approx 17 m

The sound travels to the reflector and back: $2 d = v t$ . For a distinct echo $t \geq 0.1 s$ , so

d_{m i n} = \frac{v \times 0.1}{2} \approx \frac{340 \times 0.1}{2} = 17 m

The same principle is used in SONAR (depth of the sea), in bats' echolocation and in ultrasound scanning.

Figure 12: SONAR (sound navigation and ranging) uses echoes of ultrasound:

d = \frac{v t}{2}

. With

v = 1530 m s^{- 1}

in sea water and

t = 1.2 s

d = 918 m

. Bats and ultrasound scanners use the same echo principle.

Key idea

Echo distance: $d = v t /2$ . Never forget the factor 2: the sound makes a round trip.

Quick Recall: tap to check

Two identical sources each give

60 dB

. What do they give together?

63 dB

: intensities add, and

10 lo g_{10} 2 \approx 3 dB

An echo returns after

3 s

(

v = 340 m s^{- 1}

). How far is the reflector?

d = \frac{v t}{2} = 510 m

In the bell-jar experiment, why is the bell still seen but not heard?

Light needs no medium; sound does, and the air has been pumped out.

Figure 13: Revision map: nature, range, speed (

γ P / ρ

), factors, characteristics and echo.

9. Solved Examples

Solved Example 1

Find the speed of longitudinal waves in a thin steel rod. Young's modulus of steel is

2 \times 1 0^{11} N m^{- 2}

and its density is

7800 kg m^{- 3}

. How many times faster is this than sound in air (

331 m s^{- 1}

Solution:

$v = \frac{Y}{ρ} = \frac{2 \times 1 0 ^{11}}{7800} = 2.56 \times 1 0^{7} \approx 5.06 \times 1 0^{3} m s^{- 1}$ .

Ratio: $\frac{5064}{331} \approx 15$ .

Answer: about $5.1 km s^{- 1}$ , roughly 15 times the speed in air.

Solved Example 2

Using

P = 1.01 \times 1 0^{5} Pa

ρ = 1.293 kg m^{- 3}

and

γ = 1.40

for air at NTP, find the speed of sound from Newton's formula and from Laplace's formula. What percentage error does Newton's formula make?

Solution:

Newton: $v_{N} = \frac{P}{ρ} = \frac{1.01 \times 1 0 ^{5}}{1.293} \approx 279.5 m s^{- 1}$ .

Laplace: $v_{L} = γ v_{N} = 1.40 \times 279.5 \approx 330.7 m s^{- 1}$ .

Error: $\frac{330.7 - 279.5}{330.7} \times 100 \approx 15.5%$ .

Answer: $279.5 m s^{- 1}$ and $330.7 m s^{- 1}$ ; Newton's value is about $15%$ too low.

Solved Example 3

At what temperature will the speed of sound in air be double its value at

0^{\circ} C

Solution:

$\frac{v _{2}}{v _{1}} = \frac{T _{2}}{T _{1}} = 2$ , so $T_{2} = 4 T_{1} = 4 \times 273 = 1092 K$ .

Answer: $1092 K = 819^{\circ} C$ .

Solved Example 4

The speed of sound in air at

0^{\circ} C

331 m s^{- 1}

. Find it at

27^{\circ} C

exactly, and by the approximate rule.

Solution:

Exact: $v = 331 \frac{300}{273} \approx 347.0 m s^{- 1}$ .

Approximate: $v \approx 331 + 0.61 \times 27 \approx 347.5 m s^{- 1}$ .

Answer: about $347 m s^{- 1}$ ; the rule of thumb is off by only $0.5 m s^{- 1}$ here.

Solved Example 5

The ratio of the speed of sound in hydrogen to that in oxygen at the same temperature is
(A)

1 : 4

(B)

4 : 1

(C)

16 : 1

(D)

1 : 16

Solution:

Both are diatomic (same $γ$ ), so $v \propto \frac{1}{M}$ : $\frac{v _{H_{2}}}{v _{O_{2}}} = \frac{32}{2} = 4$ .

Answer: (B).

Solved Example 6

The pressure of a gas is doubled while its temperature is kept constant. The speed of sound in it
(A) doubles
(B) becomes

2

times
(C) stays the same
(D) halves

Solution:

At constant $T$ , doubling $P$ doubles $ρ$ , so $\frac{γ P}{ρ}$ and hence $v$ are unchanged.

Answer: (C).

Solved Example 7

Estimate the ratio of the speed of sound in helium (

γ = 5/3

M = 4 g mol^{- 1}

) to that in air (

γ = 1.40

M = 29 g mol^{- 1}

) at the same temperature.

Solution:

$\frac{v _{He}}{v _{air}} = \frac{γ _{He} / M _{He}}{γ _{air} / M _{air}} = \frac{( 5/3 ) /4}{1.40/29} = 8.63 \approx 2.94$ .

Answer: about $2.9$ (NCERT values: $965/331 \approx 2.92$ ). This is why your voice sounds high after inhaling helium: the air column in the throat resonates at higher frequencies.

Solved Example 8

The speed of sound in a gas at NTP (

P = 1.01 \times 1 0^{5} Pa

ρ = 1.29 kg m^{- 3}

) is measured as

330 m s^{- 1}

. Find

γ

and say whether the gas is monatomic or diatomic.

Solution:

$γ = \frac{v ^{2} ρ}{P} = \frac{( 330 ) ^{2} \times 1.29}{1.01 \times 1 0 ^{5}} \approx 1.39$ .

Answer: $γ \approx 1.4$ : a diatomic gas (like air).

Solved Example 9

(a) By how many decibels does the sound level rise when the intensity becomes 100 times? (b) What is the intensity of a

60 dB

sound?

Solution:

(a) $Δ β = 10 lo g_{10} 100 = 20 dB$ .

(b) $60 = 10 lo g_{10} \frac{I}{1 0 ^{- 12}}$ , so $\frac{I}{1 0 ^{- 12}} = 1 0^{6}$ and $I = 1 0^{- 6} W m^{- 2}$ .

Answer: (a) $20 dB$ ; (b) $1 0^{- 6} W m^{- 2}$ .

Solved Example 10

One machine produces a sound level of

70 dB

at a point. What is the level when two identical machines run together?

Solution:

Intensities add, not decibels: $I = 2 \times 1 0^{- 5} W m^{- 2}$ (each $70 dB$ source gives $1 0^{- 5} W m^{- 2}$ ).

$β = 70 + 10 lo g_{10} 2 = 70 + 3.0 = 73 dB$ .

Answer: $73 dB$ .

Solved Example 11

A small source radiates

3.14 W

of sound power equally in all directions. Find the intensity and the sound level at

10 m

Solution:

$I = \frac{P}{4 π r ^{2}} = \frac{3.14}{4 π \times 100} = 2.5 \times 1 0^{- 3} W m^{- 2}$ .

$β = 10 lo g_{10} \frac{2.5 \times 1 0 ^{- 3}}{1 0 ^{- 12}} = 10 lo g_{10} (2.5 \times 1 0^{9}) \approx 94 dB$ .

Answer: $2.5 \times 1 0^{- 3} W m^{- 2}$ , about $94 dB$ .

Solved Example 12

A person claps near a cliff and hears the echo

2 s

later. The speed of sound is

340 m s^{- 1}

. How far is the cliff? What is the least distance at which an echo can be heard?

Solution:

$2 d = v t$ , so $d = \frac{340 \times 2}{2} = 340 m$ .

Least distance: $t = 0.1 s$ , $d_{m i n} = \frac{340 \times 0.1}{2} = 17 m$ .

Answer: $340 m$ ; $17 m$ .

Solved Example 13

Sound travels

1 km

in still air at

340 m s^{- 1}

. A wind of

20 m s^{- 1}

blows along the line of travel. Find the travel time with the wind and against it.

Solution:

With the wind: $v = 340 + 20 = 360 m s^{- 1}$ , $t = \frac{1000}{360} \approx 2.78 s$ .

Against the wind: $v = 340 - 20 = 320 m s^{- 1}$ , $t = \frac{1000}{320} \approx 3.13 s$ .

Answer: $2.78 s$ and $3.13 s$ .

Practice Questions

Find the speed of sound in water, given $B = 2.2 \times 1 0^{9} Pa$ and $ρ = 1000 kg m^{- 3}$ .Answer: $2.2 \times 1 0^{6} \approx 1483 m s^{- 1}$ .
The speed of sound in air is $330 m s^{- 1}$ at $27^{\circ} C$ . Find it at $127^{\circ} C$ .Answer: $330 400/300 \approx 381 m s^{- 1}$ .
Why is the speed of sound in hydrogen greater than in air at the same temperature?Answer: Hydrogen's molar mass ( $2$ ) is much smaller than air's ( $29$ ) and both are diatomic, so $v \propto 1/ M$ is larger.
The intensity of a sound falls to $\frac{1}{1000}$ of its value. Find the change in sound level.Answer: $- 30 dB$ .
How much louder in decibels is a point source when you move from $20 m$ to $10 m$ away?Answer: $I$ becomes 4 times: $+ 6 dB$ .
A ship's SONAR pulse returns from the sea bed after $1.2 s$ . The speed of sound in sea water is $1530 m s^{- 1}$ . Find the depth.Answer: $1530 \times 1.2/2 = 918 m$ .
Show that $v_{t} \approx v_{0} (1 + t /546)$ follows from $v \propto T$ .Answer: $v_{t} = v_{0} (1 + t /273)^{1/2} \approx v_{0} (1 + t /546)$ by the binomial approximation for small $t$ .

Common Mistakes to Avoid

Watch out

Using Newton's $v = P / ρ$ for sound in a gas. Sound is adiabatic: use $v = γ P / ρ$ .
Putting temperature in $^{\circ} C$ into $v \propto T$ . Always use kelvin.
Saying the speed of sound increases with pressure. At constant temperature it does not change.
Adding decibels directly: two $70 dB$ sources give $73 dB$ , not $140 dB$ . Add intensities, then convert.
Forgetting the round trip in echo problems: $d = v t /2$ , not $v t$ .
Thinking higher frequency sound travels faster. Speed depends on the medium; frequency changes only $λ$ .
Mixing up $Y$ and $B$ : a thin solid rod uses Young's modulus $Y$ ; liquids and gases use the bulk modulus $B$ .
Using $γ = 5/3$ for air. Air is diatomic: $γ = 7/5 = 1.40$ .

Frequently Asked Questions

Why is sound a longitudinal wave in air?

Air, like any gas, has no rigidity, so one layer cannot drag its neighbour sideways. It can only push and pull it along the direction of travel by changing its pressure. The disturbance therefore travels as compressions and rarefactions, with particles moving parallel to the wave, which is a longitudinal wave.

What is Laplace's correction to Newton's formula?

Newton treated sound in a gas as isothermal and got $v = P / ρ$ , about 280 m/s in air. Laplace showed the compressions and rarefactions are too fast for heat to flow, so the process is adiabatic, the bulk modulus is $γ P$ , and $v = γ P / ρ$ , about 331 m/s, matching experiment.

Why does sound travel faster in solids than in gases?

Speed is the square root of an elastic modulus divided by density. Solids are denser than gases, but their elastic moduli are larger by a far greater factor, about a million times for steel compared with air. So sound travels at about 5000 to 6000 m/s in metals and only about 330 m/s in air.

Does the speed of sound depend on pressure?

Not at constant temperature. When the pressure of a gas rises, its density rises in the same proportion, so the ratio of pressure to density, and hence the speed of sound, stays the same. Temperature, humidity, molar mass and wind do change the speed of sound.

How does temperature affect the speed of sound?

For a given gas the speed of sound is proportional to the square root of absolute temperature, $v \propto T$ . Near room temperature air gains about 0.61 m/s for each degree Celsius rise, so sound travels faster on a hot day than on a cold one.

What is the difference between pitch, loudness and quality of sound?

Pitch depends on frequency and tells high notes from low ones. Loudness depends on intensity and is measured in decibels. Quality, or timbre, depends on the waveform, that is the mix of overtones, and lets us tell two instruments apart even when they play the same note equally loudly.

Which sound wave topics are asked in JEE Main?

JEE Main often asks for the speed of sound with Laplace's formula, its dependence on temperature, molar mass and $γ$ , comparisons between gases, decibel calculations with intensity ratios and inverse-square spreading, and echo distances. Remember $v = γ R T / M$ and $β = 10 lo g_{10} (I / I_{0})$ .

What should NEET students learn from sound waves?

For NEET, learn why sound needs a medium, the audible range of 20 Hz to 20 kHz, Newton's formula and Laplace's correction, the effect of temperature, pressure and humidity on the speed of sound, the link between pitch and frequency, and the minimum distance of about 17 m for an echo.

Previous year questions on Sound Waves

3 questions from past papers, each with a step-by-step solution.

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